hdoj-5120-Intersection【数论】

Intersection

Time Limit: 4000/4000 MS (Java/Others) Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 964 Accepted Submission(s): 368

Problem Description

Matt is a big fan of logo design. Recently he falls in love with logo made up by rings. The following figures are some famous examples you may know.


A ring is a 2-D figure bounded by two circles sharing the common center. The radius for these circles are denoted by r and R (r < R). For more details, refer to the gray part in the illustration below.


Matt just designed a new logo consisting of two rings with the same size in the 2-D plane. For his interests, Matt would like to know the area of the intersection of these two rings.

Input

The first line contains only one integer T (T ≤ 10⁵), which indicates the number of test cases. For each test case, the first line contains two integers r, R (0 ≤ r < R ≤ 10).

Each of the following two lines contains two integers x_i, y_i (0 ≤ x_i, y_i ≤ 20) indicating the coordinates of the center of each ring.

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y is the area of intersection rounded to 6 decimal places.

Sample Input

22 30 00 02 30 05 0

Sample Output

Case #1: 15.707963Case #2: 2.250778

Source

2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

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注意： Pi 不能自己写，要用acos（-1.0） ，在这里WA了好多次

#include<stdio.h>#include<math.h>#include<algorithm>using namespace std;double dis,R,r;struct point{double x,y;}a,b;const double pi=acos(-1.0);double S(double r1,double r2){if(dis>=(r1+r2)) return 0; //外切或不相交 double s1,s2,sit1,sit2; if(dis<=fabs(r1-r2)){  //内切或内含 r1=min(r1,r2);    return pi*r1*r1; }s1=(r1*r1+dis*dis-r2*r2)/(2*r1*dis);s2=(r2*r2+dis*dis-r1*r1)/(2*r2*dis);    sit1=acos(s1);    sit2=acos(s2);return sit1*r1*r1+sit2*r2*r2-r1*dis*sin(sit1); }int main(){int t,cas=0;scanf("%d",&t);while(t--){++cas;scanf("%lf%lf",&r,&R);scanf("%lf%lf",&a.x,&a.y);scanf("%lf%lf",&b.x,&b.y);    dis=sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));    if(fabs(dis)<1e-7){    printf("Case #%d: %.6lf\n",cas,pi*(R*R-r*r));    continue;    }double ss=0;ss=S(R,R);ss-=2*S(R,r)-S(r,r);printf("Case #%d: %.6lf\n",cas,ss);}return 0;}